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Who will shout first ?

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who will shout first
Four criminals are caught in a minor offence, they pleaded judge to release them without any punishment. As it was a minor offence, judge told them that they will be freed without punishment, but only if they can solve a puzzle

All four of them are lined up on some steps(as shown in the picture above). A wall separates the 4th person from other three.
So to summarize:-
Criminal 1 can see 2 and 3
Criminal 2 can see 3
Criminal 3 and 4 can’t see any one else

Criminals are wearing hats, they are told that there are two white hats and two black hats. They can’t see their own hat color and are told to shout their hat color as soon as they get to know for certain what color it is.

They are not allowed to turn around or move.
They are not allowed to talk to each other.
They are not allowed to take their hat off.

Who will shout first ?

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January 100, February 233

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Number Puzzle #138
This is also not as much as a tough one. Let us see who can do it.
Here is a list showing the month and a number for each month .

January 100
February 233
March. 334
April. 460
May. 512
June. 645
July. 760
August. 823

Decipher the logic and find the number for September October November & December = ?

This is O P E N for all

Please support yöür answer with the logic , it’s compulsory.

When is Cheryl’s Birthday ?

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Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

May 15 May 16 May 19

June 17 June 18

July 14 July 16

August 14 August 15 August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.

Bernard: At first I didn’t know when Cheryl’s birthday is, but I know now.

Albert: Then I also know when Cheryl’s birthday is.

So when is Cheryl’s birthday?

Check your answer:-

Tried enough already?
Click here to See Answer with Solution
From the first statement, any of the day’s in that month should not be unique, else otherwise Albert can’t be sure of Bernard not knowing the birthday.

Lets take the possible answers

If it was May => Albert cant say the first statement, as if the day was 19, Bernard will know the birthday for sure.
If, it is was June => Same is the case with 18th.
If, it is in July => Albert can say the first statement.
If it is in August => Albert can say the first statement.

Now after hearing first statement, Bernard can clearly figure out that the month of birthday is either July or August.

So, if the day was 14 => It can be July 14 or August 14 =>as bernard know now, it is not possible
If the day was 15 => It can be Aug 15
If the day was 16 => it can be July 16
If the day was 17 => it can be Aug 17

Only three days possible are Aug 15, July 16 or Aug 17, bernard knows the day so he know the birthday.

From the third statement, Albert knows the birthday as well, so it can only be July 16, as if the month was Aug, Albert can not figure it out.

Thus the Cheryl’s birthday is on July 16.

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